यदि ` y=a cos (log_e x )+ b sin (log _e x ) ` तब सिद्ध कीजिए की
` x^(2) y_2 +xy_1 +y=0`

Explanation

`y=a cos (log _e x )+ b sin (log _e x ) " "...(1)`
` rArr " "y_1 =a(d)/(dx) cos (log_e x )+ b(d)/(dx) sin (log _ex)`
` =-(asin (log _e x))/(x )+(bcos log _e nx )/(x )`
` rArr " "xy_1 =- asin (log _e x )+ bcos (log _e x )`
पुनः x के सापेक्ष अवकलन करने पर
` xy_2 +y_1 =-(acos (log _e x ))/(x ) -(bsin log _e x )/(x )`
`rArr x^(2)y_2 +xy_1 =-[acos (log _e x )+ b sin (log _e x )]`
` " "=-y` [`because` समी० (1 ) से]
` therefore " "x^(2)y_2 +xy_1 +y=0`